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%I #36 Feb 12 2021 17:47:22
%S 1,6,18,24,28,40,54,72,84,96,117,120,135,162,196,200,216,224,234,252,
%T 270,288,360,384,468,486,496,540,588,600,640,648,672,756,775,819,864,
%U 891,936,1000,1080,1152,1350,1372,1458,1488,1521,1536,1550,1568,1638,1701,1764
%N Numbers k such that k divides phi(k) * sigma(k).
%C Comments from _Farideh Firoozbakht_, Dec 01 2005: (Start)
%C I. All numbers of the form 2^(4m-1)*5^n where m & n are natural numbers are in the sequence. Because if s=2^(4m-1)*5^n then phi(s)=2^(4m-2)*4*5^(n-1); sigma(s)=(2^(4m)-1)*(5^(n+1)-1)/4 so phi(s)*sigma(s)=6*((16^m-1)/15)*((5^(n+1)-1)/4)*(2^(4m-1)*5^n)= 6*((16^m-1)/15)*((5^(n+1)-1)/4)*s, note that (16^m-1)/15 and (5^(n+1)-1)/4 are integers, hence s divides phi(s)*sigma(s).
%C II. All numbers of the form 2^(2m-1)*3^n where m & n are natural numbers (A228104) are in the sequence. Because if s=2^(2m-1)*3^n then phi(s)=2^(2m-2)*2*3^(n-1); sigma(s)=(2^(2m)-1)*(3^(n+1)-1)/2 so phi(s)*sigma(s)=((3^(n+1)-1)/2)*((4^m-1)/3)*(2^(2m-1)*3^n) =((3^(n+1)-1)/2)*((4^m-1)/3)*s, note that ((3^(n+1)-1)/2 and (4^m-1)/3 are integers, hence s divides phi(s)*sigma(s).
%C So this sequence is infinite. Also it is obvious that perfect numbers (A000396) and multiply-perfect numbers(A007691) are subsequences of this sequence. (End)
%H Donovan Johnson, <a href="/A011775/b011775.txt">Table of n, a(n) for n = 1..10000</a>
%H Richard K. Guy, <a href="http://www.jstor.org/stable/2974586">Divisors and desires</a>, Amer. Math. Monthly, 104 (1997), 359-360.
%t Select[Range[1770], IntegerQ[DivisorSigma[1, # ]*EulerPhi[ # ]/# ] &] (* _Farideh Firoozbakht_, Dec 01 2005 *)
%o (PARI) is(n)=sigma(n)*eulerphi(n)%n==0 \\ _Charles R Greathouse IV_, Nov 27 2013
%Y Cf. A062354, A000396, A007691.
%K nonn
%O 1,2
%A _R. K. Guy_
%E Corrected and extended by _David W. Wilson_
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